Molecular Biology And Genetics — Part 1 MCQs with Explanations
Practice high-yield Biotechnology Engineering MCQs on Molecular Biology And Genetics — Part 1 with worked academic solutions for PPSC, FPSC, NIH, NTS.
Questions
40
Difficulty Level
Exam-Level
Target Exams
PPSC, FPSC, NIH...
Estimated Time
48 mins
Exam-Level PPSC FPSC NIH NTS
Verified for Commission Syllabus • Oct 14, 2026
40
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Question 1 of 40
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RNA is very much susceptible to hydrolysis in alkali because
Correct Answer: Option B — its 2'-OH group participate in intramolecular cleavage of phosphodiester backbone
Detailed Academic Explanation:
its 2'-OH group participate in intramolecular cleavage of phosphodiester backbone is the correct answer to “RNA is very much susceptible to hydrolysis in alkali because”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 2 of 40
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Rho-dependent and rho-independent transcription termination mechanisms operate in prokaryotes. Rho independent termination mechanism involves
Correct Answer: Option B — No protein factors and only RNA secondary structure and run of 'U's
Detailed Academic Explanation:
No protein factors and only RNA secondary structure and run of 'U's is the correct answer to “Rho-dependent and rho-independent transcription termination mechanisms operate in prokaryotes. Rho independent termination mechanism involves”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 3 of 40
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Determine the correctness or otherwise of the following Assertion (A) and the Reason (R) . Assertion: Cytoplasmic male sterility (cms) is invariably due to defect(s) in mitochondrial function. Reason: cms can be overcome by pollinating a fertility restoring (Rf ) plant with pollen from a non cms plant.
Correct Answer: Option A — Both (A) and (R) are true and (R) is the correct reason for (A)
Detailed Academic Explanation:
Both (A) and (R) are true and (R) is the correct reason for (A) is the correct answer to “Determine the correctness or otherwise of the following Assertion (A) and the Reason (R) . Assertion: Cytoplasmic male sterility (cms) is invariably due to defect(s) in mitochondrial function. Reason: cms can be overcome by pollinating a fertility restoring (Rf ) plant with pollen from a non cms plant.”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 4 of 40
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Mobile genetic elements present in human genome are P. Long interspersed elements (LINEs) Q. Short interspersed elements (SINEs) R. P elements S. IS elements
Correct Answer: Option B — P, Q
Detailed Academic Explanation:
P, Q is the correct answer to “Mobile genetic elements present in human genome are P. Long interspersed elements (LINEs) Q. Short interspersed elements (SINEs) R. P elements S. IS elements”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 5 of 40
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The direction of shell coiling in the snail Limnaea peregra is a classic example of
Correct Answer: Option B — Extra-chromosomal inheritance
Detailed Academic Explanation:
Extra-chromosomal inheritance is the correct answer to “The direction of shell coiling in the snail Limnaea peregra is a classic example of”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 6 of 40
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Which of the following characteristics with respect to bacterial DNA polymerase III are true? P. Initiation of chain synthesis Q. 5' → 3' polymerization R. 3' → 5' exonuclease activity S. 5' → 3' exonuclease activity
Correct Answer: Option B — Q and R only
Detailed Academic Explanation:
Q and R only is the correct answer to “Which of the following characteristics with respect to bacterial DNA polymerase III are true? P. Initiation of chain synthesis Q. 5' → 3' polymerization R. 3' → 5' exonuclease activity S. 5' → 3' exonuclease activity”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 7 of 40
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The protein in eukaryotes which is subjected to degradation undergoes
Correct Answer: Option C — Ubiquitation
Detailed Academic Explanation:
Ubiquitation is the correct answer to “The protein in eukaryotes which is subjected to degradation undergoes”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 8 of 40
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In mismatch correction repair, the parental DNA strand is distinguished from the daughter strand by
Correct Answer: Option C — Methylation
Detailed Academic Explanation:
Methylation is the correct answer to “In mismatch correction repair, the parental DNA strand is distinguished from the daughter strand by”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 9 of 40
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Which one of the following is coded by the ABO blood group locus in the human genome?
Correct Answer: Option B — Galactosyltransferase
Detailed Academic Explanation:
Galactosyltransferase is the correct answer to “Which one of the following is coded by the ABO blood group locus in the human genome”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 10 of 40
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Determine the correctness or otherwise of the following Assertion (A) and the Reason (R) . Assertion: The association constant in water for the G-C base pair is three times lower than that for the A-T base pair. Reason: There are three hydrogen bonds in the G-C base pair and two in the A-T base pair.
Correct Answer: Option B — (A) is false, but (R) is true
Detailed Academic Explanation:
(A) is false, but (R) is true is the correct answer to “Determine the correctness or otherwise of the following Assertion (A) and the Reason (R) . Assertion: The association constant in water for the G-C base pair is three times lower than that for the A-T base pair. Reason: There are three hydrogen bonds in the G-C base pair and two in the A-T base pair.”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 11 of 40
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5' capping of mRNA transcripts in eukaryotes involves, the following events: P. Addition of GMP on the 5' end. Q. Removal of γ-phosphate of the triphosphate on first base at the 5' end. R. 5' - 5' linkage between GMP and the first base at 5' end. S. Addition of methyl group to N7 position of guanine. Which one of the following is the correct sequence of events?
Correct Answer: Option C — Q, P, R, S
Detailed Academic Explanation:
Q, P, R, S is the correct answer to “5' capping of mRNA transcripts in eukaryotes involves, the following events: P. Addition of GMP on the 5' end. Q. Removal of γ-phosphate of the triphosphate on first base at the 5' end. R. 5' - 5' linkage between GMP and the first base at 5' end. S. Addition of methyl group to N7 position of guanine. Which one of the following is the correct sequence of events”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 12 of 40
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Determine the correctness of the following statements. I. Enhancer sequences are those DNA sequences that are involved in increasing the rate of DNA replication. II. Enhancer sequences work by binding with eukaryotic gene activator factors.
Correct Answer: Option B — Only II is true
Detailed Academic Explanation:
Only II is true is the correct answer to “Determine the correctness of the following statements. I. Enhancer sequences are those DNA sequences that are involved in increasing the rate of DNA replication. II. Enhancer sequences work by binding with eukaryotic gene activator factors.”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 13 of 40
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Based on their function, find the odd one out
Correct Answer: Option D — snRNA
Detailed Academic Explanation:
snRNA is the correct answer to “Based on their function, find the odd one out”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 14 of 40
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For their efficient translation, eubacterial mRNAs possess a Shine-Dalgarno sequence for its recognition by an anti-Shine-Dalgarno sequence (ASD) in the ribosomes. The correct statement is
Correct Answer: Option C — ASD is present in 16S rRNA
Detailed Academic Explanation:
ASD is present in 16S rRNA is the correct answer to “For their efficient translation, eubacterial mRNAs possess a Shine-Dalgarno sequence for its recognition by an anti-Shine-Dalgarno sequence (ASD) in the ribosomes. The correct statement is”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 15 of 40
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In transgenics, alterations in the sequence of nucleotide in genes are due to P. Substitution Q. Deletion R. Insertion S. Rearrangement
Correct Answer: Option B — P, Q and R
Detailed Academic Explanation:
P, Q and R is the correct answer to “In transgenics, alterations in the sequence of nucleotide in genes are due to P. Substitution Q. Deletion R. Insertion S. Rearrangement”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 16 of 40
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Peptidyl transferase activity resides in
Correct Answer: Option B — 23S rRNA
Detailed Academic Explanation:
23S rRNA is the correct answer to “Peptidyl transferase activity resides in”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 17 of 40
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A DNA sequence, 5'-ATGGACGTGCTTCCCAAAG-CATCGGGC-3', is mutated to obtain P. 5'- ATGGACGTGCTTCaCAAAGCATCGGGC -3' Q. 5'- ATGGACGTGCTTCCCgAAAGCATCGGGC -3' R. 5'- ATGGACGTGCTTCC-AAAGCATCGGGC -3' S. 5'- ATGGACGTGCTTCCCAAtGCATCGGGC -3' T. 5'- ATGGACGaGCTTCCCAAAGCATCGGGC -3' [Point mutations are shown in the lower case or '-' within the sequences] Which of the above mutant sequences do not have frameshift?
Correct Answer: Option B — P, S and T
Detailed Academic Explanation:
P, S and T is the correct answer to “A DNA sequence, 5'-ATGGACGTGCTTCCCAAAG-CATCGGGC-3', is mutated to obtain P. 5'- ATGGACGTGCTTCaCAAAGCATCGGGC -3' Q. 5'- ATGGACGTGCTTCCCgAAAGCATCGGGC -3' R. 5'- ATGGACGTGCTTCC-AAAGCATCGGGC -3' S. 5'- ATGGACGTGCTTCCCAAtGCATCGGGC -3' T. 5'- ATGGACGaGCTTCCCAAAGCATCGGGC -3' [Point mutations are shown in the lower case or '-' within the sequences] Which of the above mutant sequences do not have frameshift”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 18 of 40
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DNA helicases catalyze the reaction
Correct Answer: Option B — DNA unwinding
Detailed Academic Explanation:
DNA unwinding is the correct answer to “DNA helicases catalyze the reaction”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 19 of 40
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The formation of peptide cross links between adjacent glycan chains in cell wall synthesis is called
Correct Answer: Option D — Transpeptidation
Detailed Academic Explanation:
Transpeptidation is the correct answer to “The formation of peptide cross links between adjacent glycan chains in cell wall synthesis is called”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 20 of 40
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In a relatively large but finite and closed population of sexually reproducing diploid organisms, the frequency of homozygous genotype PP changes from 0.40 to 0.50 and that of pp changes from 0.40 to 0.41 in a span of 10 generations. Which of the following is the most likely cause for the above change in frequency of the PP genotype?
Correct Answer: Option C — Selection
Detailed Academic Explanation:
Selection is the correct answer to “In a relatively large but finite and closed population of sexually reproducing diploid organisms, the frequency of homozygous genotype PP changes from 0.40 to 0.50 and that of pp changes from 0.40 to 0.41 in a span of 10 generations. Which of the following is the most likely cause for the above change in frequency of the PP genotype”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 21 of 40
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A protein is phosphorylated at a serine residue. A phosphomimic mutant of the protein can be generated by substituting that serine with
Correct Answer: Option C — Aspartate
Detailed Academic Explanation:
Aspartate is the correct answer to “A protein is phosphorylated at a serine residue. A phosphomimic mutant of the protein can be generated by substituting that serine with”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 22 of 40
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Under alkaline conditions, DNA is more stable than RNA because
Correct Answer: Option D — RNA is susceptible to hydrolysis
Detailed Academic Explanation:
RNA is susceptible to hydrolysis is the correct answer to “Under alkaline conditions, DNA is more stable than RNA because”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 23 of 40
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True breeding Drosophila flies with curved wings and dark bodies were mated with true breeding short wings and tan body Drosophila, The F 1 progeny was observed to be with curved wings and tan body. The F 1 progeny was again allowed to breed and produced flies of the following phenotype, 45 curved wings tan body, 15 short wings tan body, 16 curved wings dark body and, 6 short wings dark body. The mode of inheritance is
Correct Answer: Option A — Typical Mendelian with curved wings and tan body being dominant
Detailed Academic Explanation:
Typical Mendelian with curved wings and tan body being dominant is the correct answer to “True breeding Drosophila flies with curved wings and dark bodies were mated with true breeding short wings and tan body Drosophila, The F 1 progeny was observed to be with curved wings and tan body. The F 1 progeny was again allowed to breed and produced flies of the following phenotype, 45 curved wings tan body, 15 short wings tan body, 16 curved wings dark body and, 6 short wings dark body. The mode of inheritance is”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 24 of 40
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The enzyme which adds phosphate group to the free 5' terminus of a DNA sequence is
Correct Answer: Option C — Polynucleotide kinase
Detailed Academic Explanation:
Polynucleotide kinase is the correct answer to “The enzyme which adds phosphate group to the free 5' terminus of a DNA sequence is”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 25 of 40
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The fundamental feature of the genetic code which allows the expression of a protein in any host is its
Correct Answer: Option B — Universality
Detailed Academic Explanation:
Universality is the correct answer to “The fundamental feature of the genetic code which allows the expression of a protein in any host is its”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 26 of 40
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Match the terms in Group I with their possible explanation in Group II Group I Group II P. Orthologs 1. A cell or an organism, having foreign gene Q. Paralogs 2. The complement of a protein expressed by a genome R. Proteome 3. Genes from different species related to each other S. Transgenic 4. Genes from same species related to each other
Correct Answer: Option B — P-4, Q-3, R-2, S-1
Detailed Academic Explanation:
P-4, Q-3, R-2, S-1 is the correct answer to “Match the terms in Group I with their possible explanation in Group II Group I Group II P. Orthologs 1. A cell or an organism, having foreign gene Q. Paralogs 2. The complement of a protein expressed by a genome R. Proteome 3. Genes from different species related to each other S. Transgenic 4. Genes from same species related to each other”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 27 of 40
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Which one of the following amino acids in proteins does not undergo phosphorylation?
Correct Answer: Option C — Pro
Detailed Academic Explanation:
Pro is the correct answer to “Which one of the following amino acids in proteins does not undergo phosphorylation”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 28 of 40
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In eukaryotes, the ribosomal RNA genes are transcribed by
Correct Answer: Option C — RNA polymerase I
Detailed Academic Explanation:
RNA polymerase I is the correct answer to “In eukaryotes, the ribosomal RNA genes are transcribed by”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 29 of 40
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The repeat sequence of telomere in humans is
Correct Answer: Option B — 5'-TTAGGG-3'
Detailed Academic Explanation:
5'-TTAGGG-3' is the correct answer to “The repeat sequence of telomere in humans is”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 30 of 40
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Match the terms in Group I with their associated functions in Group II . Group I Group II P. Shine-Dalgarno sequences 1. Aminoacylation of tRNA Q. Leucine zipper 2. Gene silencing R. Aminoacyl tRNA synthetase 3. Ribosome binding and facilitation of translation initiation S. RNA interference (RNAi) 4. Transcription factors
Correct Answer: Option A — P-3, Q-4, R-1, S-2
Detailed Academic Explanation:
P-3, Q-4, R-1, S-2 is the correct answer to “Match the terms in Group I with their associated functions in Group II . Group I Group II P. Shine-Dalgarno sequences 1. Aminoacylation of tRNA Q. Leucine zipper 2. Gene silencing R. Aminoacyl tRNA synthetase 3. Ribosome binding and facilitation of translation initiation S. RNA interference (RNAi) 4. Transcription factors”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 31 of 40
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In nature, the horizontal gene transfer across bacteria is mediated by
Correct Answer: Option B — Conjugation and transformation
Detailed Academic Explanation:
Conjugation and transformation is the correct answer to “In nature, the horizontal gene transfer across bacteria is mediated by”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 32 of 40
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Which one of the following covalent linkages exists between 7-Methyl guanosine (m7G) and mRNAs?
Correct Answer: Option C — 5' - 5' triphosphate
Detailed Academic Explanation:
5' - 5' triphosphate is the correct answer to “Which one of the following covalent linkages exists between 7-Methyl guanosine (m7G) and mRNAs”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 33 of 40
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siRNA(s) interfere at
Correct Answer: Option B — Post-transcriptional level
Detailed Academic Explanation:
Post-transcriptional level is the correct answer to “siRNA(s) interfere at”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 34 of 40
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Determine the correctness or otherwise of the following Assertion (A) and the Reason (R) . Assertion: Isopropylthiogalactoside (IPTG) is a gratuitous inducer of lactose operon. Reason: Gratuitous inducers are chemical analogs which behave like natural inducer but they do not serve as substrate for the enzymes that are subsequently synthesized.
Correct Answer: Option A — Both (A) and (R) are true and (R) is the correct reason for (A)
Detailed Academic Explanation:
Both (A) and (R) are true and (R) is the correct reason for (A) is the correct answer to “Determine the correctness or otherwise of the following Assertion (A) and the Reason (R) . Assertion: Isopropylthiogalactoside (IPTG) is a gratuitous inducer of lactose operon. Reason: Gratuitous inducers are chemical analogs which behave like natural inducer but they do not serve as substrate for the enzymes that are subsequently synthesized.”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 35 of 40
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Hypophosphatemia is manifested by an X-linked dominant allele. What proportion of the offspring from a normal male and an affected heterozygous female will manifest the disease?
Correct Answer: Option A — $${1 \over 2}$$ sons and $${1 \over 2}$$ daughters
Detailed Academic Explanation:
$${1 \over 2}$$ sons and $${1 \over 2}$$ daughters is the correct answer to “Hypophosphatemia is manifested by an X-linked dominant allele. What proportion of the offspring from a normal male and an affected heterozygous female will manifest the disease”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 36 of 40
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Match the classes of RNA molecules in Group I with their functions in Group II . Group I Group II P. snoRNA 1. Protects germline from transposable elements Q. piRNA 2. Blocks translation of selected mRNA R. miRNA 3. Template for telomere elongation S. snRNA 4. Modification and processing of rRNA 5. Splicing of RNA transcripts
Correct Answer: Option D — P-4, Q-1, R-2, S-5
Detailed Academic Explanation:
P-4, Q-1, R-2, S-5 is the correct answer to “Match the classes of RNA molecules in Group I with their functions in Group II . Group I Group II P. snoRNA 1. Protects germline from transposable elements Q. piRNA 2. Blocks translation of selected mRNA R. miRNA 3. Template for telomere elongation S. snRNA 4. Modification and processing of rRNA 5. Splicing of RNA transcripts”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 37 of 40
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Reverse genetics means
Correct Answer: Option A — Finding the function of an ORF
Detailed Academic Explanation:
Finding the function of an ORF is the correct answer to “Reverse genetics means”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 38 of 40
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The cellular process which utilizes RNA-induced silencing complex to block gene expression is
Correct Answer: Option B — RNA interference
Detailed Academic Explanation:
RNA interference is the correct answer to “The cellular process which utilizes RNA-induced silencing complex to block gene expression is”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 39 of 40
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Match the type of chromosomal inheritance (Column I) with the corresponding genetic disease or trait (Column II) . Column I Column II P. Autosomal recessive inheritance 1. Huntington disease Q. Autosomal dominant inheritance 2. Hairy ears R. X-linked inheritance 3. Cystic fibrosis S. Y-linked inheritance 4. Hemophilia
Correct Answer: Option C — P-3, Q-1, R-4, S-2
Detailed Academic Explanation:
P-3, Q-1, R-4, S-2 is the correct answer to “Match the type of chromosomal inheritance (Column I) with the corresponding genetic disease or trait (Column II) . Column I Column II P. Autosomal recessive inheritance 1. Huntington disease Q. Autosomal dominant inheritance 2. Hairy ears R. X-linked inheritance 3. Cystic fibrosis S. Y-linked inheritance 4. Hemophilia”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 40 of 40
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Protein P becomes functional upon phosphorylation of a serine residue, replacing this serine with . . . . . . . . will result in a phosphomimic mutant of P.
Correct Answer: Option B — Aspartic acid
Detailed Academic Explanation:
Aspartic acid is the correct answer to “Protein P becomes functional upon phosphorylation of a serine residue, replacing this serine with . . . . . . . . will result in a phosphomimic mutant of P.”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
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Analytical Tools In Biotechnology — Part 1
Master the next curriculum module with 40 exam-standard MCQs.
💡Examiner's Advisory for Biotechnology Engineering
In competitive commission exams (FPSC, PPSC, KPPSC, SPSC, CSS, PMS), questions are frequently phrased with distractor options designed to catch rote learners. Whenever you answer an MCQ, read the Detailed Academic Explanation and Conceptual Anchor to understand why the remaining options are incorrect.